In his book Young Tableaux, Fulton asserts, in Exercise 9.4.18 on p. 152, that the Schubert variety $\Omega_{\lambda}$ is defined by the conditions $\text{dim}(V \cap F_{n+i- \lambda_{i}}) \geq i$ for those $i$ such that the coordinate $(i,\lambda_{i})$ is an outside corner of the Young diagram $\lambda$. I take it that $F_{\cdot}$ is the standard flag variety, and so the first initial part of this exercise makes sense in terms of the complex Grassmannian $Gr^{n}(E)=Gr_{r}(E)$ with $E$ an $m$-dimensional complex vector space such that $m=n+r$, i.e. $$\Omega_{\lambda}= \{V \in Gr^{n}(E): \text{dim}(V \cap F_{n+i- \lambda_{i}}) \geq i,1 \leq i \leq r\}.$$ However, how does one show that each $i$ is such that $(i,\lambda_{i})$ is an outside corner of the Young diagram $\lambda$? It's difficult for me to visualize it. What are some good sources that might help to explain this better?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ The second part asks you to prove that none of the conditions can be omitted, right? Your restatement doesn't make that much sense to me. $\endgroup$– darij grinbergCommented Apr 13, 2016 at 22:11
-
$\begingroup$ Yes, but that's not the part that is giving me a hard time. $\endgroup$– LibertronCommented Apr 13, 2016 at 22:12
-
$\begingroup$ Being the ouside corners of a Young diagram just means that you have some set of pairs $(i,\lambda_i)$, no two having the same values in the first coordinate or in the second coordinate, and such that if $i_1<i_2$, then $\lambda_{i_1}>\lambda_{i_2}$ (note: the inequalities go in opposite directions). $\endgroup$– Hugh ThomasCommented Apr 14, 2016 at 2:10
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
All values of $(i,\lambda_i)$ appear. The ones that aren't outside corners are redundant.
answered Apr 14, 2016 at 1:02